Karim Belabas on Mon, 11 Nov 2024 08:51:27 +0100 |
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Re: How to calculate the conductor of an abelian extension such as Q[x]/(x^3- 19x -19) |
* David Bernier [2024-11-11 05:08]: > I'm interested in cubic extensions of Q that are abelian, in connection with > a probable prime test. I have a list of cubic polynomials f_1, ... f_22 and > I want to find the first f_i such that f_i is irreducible over F_p, where p > can be assumed prime. For a given f_i, I noticed a periodicity in p of the > irreducibility character of f_i over F_p (ref. Mathematics Stack Exchange at > the link: https://math.stackexchange.com/questions/4995484/irreducibility-of-cubic-polynomials-over-finite-fields-f-p > ). User leoli1 mentioned as relevant the conductor N of the splitting field > of f_i. I have f_8 = X^3 - 19X - 19 with discriminant 133^2. (You meant 19^2.) > How could I calculate the conductor of Q[x]/(f_8) in PARI/gp? In general (this would work for an abelian extension of an arbitrary base number field): Q = bnfinit(y); \\ = field of rational numbers f8 = x^3 - 19*x - 19; \\ assumed abelian over Q rnfconductor(Q, f8)[1] \\ or rnfconductor(Q, f8, 2)[1] the answer is [Mat(19), [0]], which is a pair [f0, foo], where - foo is the infinite part of the conductor (here trivial), - f0 is the finite part as an ideal of the maximal order (= the ring Z) of the base field (here 19 Z) However, in this particular case of a cyclic cubic number field K, it is know (cf., e.g., Hasse [1] but it is a 5 lines proof ... assuming class field theory) that - foo is always trivial - f0 is of the form f.Z where f^2 is the discriminant of K. Hence, f = sqrtint(nfdisc(f8)) gives the answer, in a rather simpler way. [1] Hasse, H., Arithmetische Theorie der kubischen Zahlkörper auf klassenkörpertheoretischer Grundlage, Math. Zeitschrift (31), 1930, pp. 565-582. Cheers, K.B. -- Pr. Karim Belabas, U. Bordeaux, Vice-président en charge du Numérique Institut de Mathématiques de Bordeaux UMR 5251 - (+33) 05 40 00 29 77 http://www.math.u-bordeaux.fr/~kbelabas/